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OLga [1]
3 years ago
12

Differential Equations - Reduction of order

Mathematics
1 answer:
iren [92.7K]3 years ago
6 0
With reduction of order, we assume a solution of the form y_2=zy_1=ze^{2x}, with z=z(x). Then

{y_2}'=(z'+2z)e^{2x}
{y_2}''=(z''+4z'+4z)e^{2x}

and substituting into the ODE gives

x(z''+4z'+4z)e^{2x}-(2x+1)(z'+2z)e^{2x}+2ze^{2x}=0
x(z''+4z'+4z)-(2x+1)(z'+2z)+2z=0
xz''+(2x-1)z'=0

Let \xi(x)=z'(x), so that \xi'=z''. This gives the linear ODE

x\xi'+(2x-1)\xi=0

This equation is also separable, so you can write

\dfrac{\xi'}{\xi}=\dfrac{1-2x}x

Integrating both sides with respect to x gives

\ln|\xi|=-2x+\ln x+C_1
\xi=C_1xe^{-2x}

Next, solve z'=\xi for z by integrating both sides again with respect to x.

z'=\xi
\implies z=\displaystyle\int C_1xe^{-2x}\,\mathrm dx
\implies z=C_1e^{-2x}(2x+1)+C_2

And finally, solve for y_2.

y_2=zy_1=C_1(2x+1)+C_2e^{2x}

and note that y_1 is already taken into account as part of y_2, so this is the general solution to the ODE.
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En una hoja de papel cuyo perímetro es de 96 centímetros, se quiere imprimir un volante de manera que el área impresa sea un rec
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El perímetro de la región impresa es 72 cm y su área es 288 cm².  

Step-by-step explanation:

1. Tenemos el perímetro de la hoja de papel:

P₁ = 96 cm = 2l₁ + 2a₁  (1)  

Como sabemos el margen superior, inferior, izquierdo y derecho podemos encontrar la relación entre el largo y ancho del rectángulo interno (región impresa) con el largo (l) y ancho (a) del rectángulo externo (hoja de papel):      

l_{2} = l_{1} - (m_{s} + m_{i}) = l_{1} - (3 cm + 2 cm) = l_{1} - 5 cm  (2)            

a_{2} = a_{1} - (m_{d} + m_{iz}) = a_{1} - (2 cm + 5 cm) = a_{1} - 7 cm   (3)    

El perímetro del rectángulo interno es:

P_{2} = 2l_{2} + 2a_{2}    (4)

Introduciendo la ecuación (2) y (3) en (4):

P_{2} = 2l_{2} + 2a_{2} = 2(l_{1} - 5 cm) + 2(a_{1} - 7 cm) = 2l_{1} + 2a_{1} - 10 cm - 14 cm = 96 cm - 24 cm = 72 cm  

Por lo tanto el perímetro del rectángulo interno (región impresa) es 72 cm.

 

2. Ahora para encontrar el área rectángulo interno debemos encontrar el largo y ancho del mismo, sabiendo que:

l_{2} = 2a_{2}     (5)

Introduciendo (5) en (4):

P_{2} = 2l_{2} + 2a_{2} = 2*2a_{2} + 2a_{2} = 6a_{2}

a_{2} = \frac{P_{2}}{6} = \frac{72 cm}{6} = 12 cm

l_{2} = 2a_{2} = 2*12 cm = 24 cm

Entonces el área es:

A_{2} = l_{2}*a_{2} = 12 cm*24 cm = 288 cm^{2}

Por lo tanto el área del rectágulo interno (región impresa) es 288 cm².      

Espero que te sea de utilidad!  

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The graph of a linear function passes through points A(-2, 5) and B(4, -4). What is the rate of change of the graph?
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When using the the slope formula the equation would be

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